Answer to Question #164466 in Chemistry for Phyroe

Question

What is the boiling point of a solution of cane sugar, C12H22O11, containing 100 g of the solute per kilogram of water?

Answer 1

ΔT_b = K_b · b_B

K_b, the ebullioscopic constant, which is dependent on the properties of the solvent (for water it is equal to 0.513 (°C·kg)/mol.

b_B is the molality of the solution, calculated by taking dissociation into account since the boiling point elevation is a colligative property, dependent on the number of particles in the solution.

The formula for molality is m = moles of solute / kilograms of solvent.

n = m / M

M (C₁₂H₂₂O₁₁) = 342.3 g/mol

n (C₁₂H₂₂O₁₁) = 100 / 342.3 = 0.3 mol

Molality m (C₁₂H₂₂O₁₁) = 0.3 / 1 = 0.3 m

ΔT_b = 0.513 · 0.3 = 0.2°

Answer: the boiling point of a solution of cane sugar, C12H22O11, containing 100 g of the solute per kilogram of water is 100.2°